TITLE: Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
QUESTION [13 upvotes]: Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$


This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
Method 2:
$$S=\text{coefficient of $x^n$ in }:$$
$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$
$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$
It looks like a heavy task to collect all the $x^n$ coefficients from this expression.

Im out of ideas. Any hint is appreciated.
From Putnam 2020, Q A2

REPLY [2 votes]: OP's second method can be made to work and it gives the simplest as in the following solution.
$$S=\text{Coefficient of $x^n$ in}~~ (1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right).$$
$$S=-\text{Coefficient of $x^n$}~ \text{in} \left((1+x)^{2n+1}(1-x)^{-1}-2^{n+1}(1-x)^{-1}\right)$$
$$S=-\text{Coefficient of $x^n$}~ \text{in} \left((1+x)^{2n+1}\sum_{k=0}^{\infty} x^k-2^{n+1}\sum_{k=0}^{\infty}x^k\right)$$
$$S=- \left(\sum_{k=0}^{n} {2n+1\choose k}-2^{n+1}\sum_{k=0}^{\infty}{n \choose k}\right)=-2^{2n}+2^{2n+1}=2^{2n}$$